General Drop Tank Tender?

[Grav_Moped]

But, that opens the door to massively overloading ships with high-density cargo, if the mass doesn't matter, you can fill that 14 cubic meter per ton cargo bay with lead, which will weigh 158.8 tons (https://www.aqua-calc.com/calculate/volume-to-weight). That seems wrong also.

True. It's a game though.

In LBB2, you're allowed +/- 10-20% leeway for deck plans, so density can vary. LBB2'77 was not entirely clear or consistent, and may in places have actually been written in metric tons rather than 13.5m³ Td (Displacement Tons). LBB1-3 '81 cleaned that up though

And IIRC, one or more of the later rule sets gave an average mass of 10,000kg/Td.

 

[Badenov]

True. It's a game though.

In LBB2, you're allowed +/- 10-20% leeway for deck plans, so density can vary. LBB2'77 was not entirely clear or consistent, and may in places have actually been written in metric tons rather than 13.5m³ Td (Displacement Tons). LBB1-3 '81 cleaned that up though

And IIRC, one or more of the later rule sets gave an average mass of 10,000kg/Td.

So 10 mass-tons per displacement ton? So are the cargoes you roll for the availability of in mass-tons or disp tons? This disconnect seems to really skew things. And thrusters work on volume rather than mass also? That seems counter-intuitive.

 

[mike wightman]

In CT the cargo ton was 1000kg by mass.

I reconcile this by having a cargo displacement ton being limited to 14 cubic metres of volume and no more than 1000kg mass.

 

[AnotherDilbert]

But, that opens the door to massively overloading ships with high-density cargo, if the mass doesn't matter, you can fill that 14 cubic meter per ton cargo bay with lead, which will weigh 158.8 tons (https://www.aqua-calc.com/calculate/volume-to-weight). That seems wrong also.

I think we can safely assume it's a simplification... Mass matters, but we can't be bothered with all the calculations.

In the editions that care, cargo is assumed to be about 1 tonne per m³. (Except LBB2'77 that was probably mass-based, not volume based, so 1 tonne per displacement ton.)

Fuel (liquid hydrogen) has very low density, so its mass isn't all that important.

 

[Badenov]

In CT the cargo ton was 1000kg by mass.

I reconcile this by having a cargo displacement ton being limited to 14 cubic metres of volume and no more than 1000kg mass.

I think we can safely assume it's a simplification... Mass matters, but we can't be bothered with all the calculations.

In the editions that care, cargo is assumed to be about 1 tonne per m³. (Except LBB2'77 that was probably mass-based, not volume based, so 1 tonne per displacement ton.)

Fuel (liquid hydrogen) has very low density, so its mass isn't all that important.

This is how we play it in my current game also: 1T/1dT of spaceship is 14m^3 volume and no more than 1000kg mass. Hydrogen gas hits both of these at roughly the same time, but for instance the abovementioned lead fills 1/158th of the volume. It's just a smidgen awkward when carrying normal solids whose densities are so much more than hydrogen.

 

[Spinward Flow]

2. An additional 2.75Td*pn (LBB2) or 0.275%/Jn of fuel (LBB5) and a power plant that can turn it into power is needed to complete the 1-week Jump with an allowance for a +10% duration variance. This is because power plant fuel burn rate for starships is the same as for non-starships; thus, the power plant must be running throughout the duration of the Jump -- and is required to do so for some reason. MgT requires a lot less than 1EP constant output during Jump -- all that's needed is basic overhead power.

This is an interesting point of divergence and actually matters with respect to fuel consumption through jumps.

1. Power Plant needs to be running @ Pn=Jn during the 1 week of jump.
2. Power Plant merely needs to provide "housekeeping" power levels during jump, since no maneuvering is possible, nor is there any reason to keep weapons powered and screens active while in jump. Computer model power needs will need to be met while in jump, however.

It's pretty clear from the way that LBB2.81 and LBB5.80 are written that jump fuel is consumed upon jumping in a single surge, rather than being something that gets consumed over the duration of the week in jump. Your 100 starship @ J2 isn't spending 20/7=2.857 tons of fuel per day while in jump (for example). You spend 20 tons of fuel to enter jump ... and then you just "coast along" until breakout.

 

[AnotherDilbert]

This is how we play it in my current game also: 1T/1dT of spaceship is 14m^3 volume and no more than 1000kg mass. Hydrogen gas hits both of these at roughly the same time, but for instance the abovementioned lead fills 1/158th of the volume. It's just a smidgen awkward when carrying normal solids whose densities are so much more than hydrogen.

Sorry, but that does not make much sense unless the standard interstellar cargo is goose down.
1 td cargo hold being 14 m³ AND 1 tonne would leave most of the cargo hold very empty, most of the time, wasting a lot of expensive ship.

You can reasonably either say:
(with LBB2'77) that 1 Dton = 1 tonne (& hence not 14 m³) or,
(with LBB2'81) that 1 Dton = 14 m³ (& hence not 1 tonne).

The obvious disadvantage with the first assumption is that you have no handy way to scale deck plans.

 

[Grav_Moped]

This is an interesting point of divergence and actually matters with respect to fuel consumption through jumps.

1. Power Plant needs to be running @ Pn=Jn during the 1 week of jump.
2. Power Plant merely needs to provide "housekeeping" power levels during jump, since no maneuvering is possible, nor is there any reason to keep weapons powered and screens active while in jump. Computer model power needs will need to be met while in jump, however.

It's pretty clear from the way that LBB2.81 and LBB5.80 are written that jump fuel is consumed upon jumping in a single surge, rather than being something that gets consumed over the duration of the week in jump. Your 100 starship @ J2 isn't spending 20/7=2.857 tons of fuel per day while in jump (for example). You spend 20 tons of fuel to enter jump ... and then you just "coast along" until breakout.

Not really -- but it is clearly how LBB2'77 works, and how MgT works -- the former just assumes basic overhead power draw is trivial while the latter details it as being far less than the equivalent of 1 (CT) Energy Point per turn. Light off the jump drive and coast.

J2 (in a Type S with the LBB2 rate) needs 20 Tons fuel to break into Jumpspace up front, and 5.5 tons (40 tons for a month +10% variance / 4 weeks = 5.5 Ton per week) fuel for the week of Jump. The 20 Tons gets burned instantaneously (ok, in either 11 or 20-40 minutes depending on the ruleset), while the 5.5 Tons are burned over the course of the next 168 (+/-10%) hours. I presume that the reason that this rate is much higher than what's needed to just keep the lights on and the air cleaners running is because the jump drive needs some juice to keep Jumpspace at bay until just before jump exit.

Again, this is CT HG'80 and LBB2'81 -- LBB2'77 and MgT don't have the additional Jump power requirement during Jump transit.

 

[Badenov]

Sorry, but that does not make much sense unless the standard interstellar cargo is goose down.
1 td cargo hold being 14 m³ AND 1 tonne would leave most of the cargo hold very empty, most of the time, wasting a lot of expensive ship.

You can reasonably either say:
(with LBB2'77) that 1 Dton = 1 tonne (& hence not 14 m³) or,
(with LBB2'81) that 1 Dton = 14 m³ (& hence not 1 tonne).

The obvious disadvantage with the first assumption is that you have no handy way to scale deck plans.

Well, yes, that makes sense. It's just never explained that way in the rules. 'Only the fuel tank is 14m^3/ton' or something would be helpful, or counting space and mass separately like in the Military Vehicles supplement. We don't normally haul lead, of course, but most of our cargoes don't fill the cargo bay floor-to-ceiling, nor do they cover the floor. They stack neatly, with place to walk between them.

 

[Spinward Flow]

Sorry, but that does not make much sense unless the standard interstellar cargo is goose down.

Let's use a slightly remarkable contemporary item then ... an F-16 ... to illustrate the point.

• Length: 15.06m
• Width: 9.96m
• Height: 4.9m
• Empty Weight: 8,573 kg (because we're shipping one as cargo)

15.06 * 9.96 * 4.9 / 14 = 52.49916 displacement tons
Mass = 8.573 tons

So if you're transporting an F-16 as cargo in a flyable condition (just as fuel and armaments) then you're going to want a 53 ton (minimum) cargo space to put it in. If you're transporting an F-16 as parts loaded into crates (some assembly required...) then you need a 9 ton (minimum) cargo mass capacity to transport it.

1 td cargo hold being 14 m3 AND 1 tonne would leave most of the cargo hold very empty, most of the time, wasting a lot of expensive ship.

More like 1000kg OR 14,000 liters, whichever limit you hit first, amounts to "1 ton" of cargo capacity for manifest spreadsheet and thus ticket purchases.

Some cargoes will be very high density (metal alloys come to mind) so you won't need all the volume available, because you're hitting the "mass limit per ton" first, as opposed to bumping up against the volume limit first. Conversely, some lower density items (finished goods that are not especially compact, such as an atmospheric fighter jet) will "waste" a lot of volume simply due to how they're shaped in comparison to the "box" that they would be fit into.

This is how you can have an F-16 fighter jet need 53 tons of displacement capacity.

And just for anyone playing the home game, the empty weight of an F-6 divided by a volume defined by the dimensions of an F-16 (length, width, height) yields an overall average density (specific gravity) of 8.5/52.5 ≈ 0.1619 ... or slightly under 1/6th the density of water, when the F-16 is "safed" for transport ... because so much of the volume defined by the aircraft's maximum dimensions is filled with "mostly empty air" when transported aboard another craft.

 

[kilemall]

Let's use a slightly remarkable contemporary item then ... an F-16 ... to illustrate the point.

• Length: 15.06m
• Width: 9.96m
• Height: 4.9m
• Empty Weight: 8,573 kg (because we're shipping one as cargo)

15.06 * 9.96 * 4.9 / 14 = 52.49916 displacement tons
Mass = 8.573 tons

So if you're transporting an F-16 as cargo in a flyable condition (just as fuel and armaments) then you're going to want a 53 ton (minimum) cargo space to put it in. If you're transporting an F-16 as parts loaded into crates (some assembly required...) then you need a 9 ton (minimum) cargo mass capacity to transport it.

More like 1000kg OR 14,000 liters, whichever limit you hit first, amounts to "1 ton" of cargo capacity for manifest spreadsheet and thus ticket purchases.

Some cargoes will be very high density (metal alloys come to mind) so you won't need all the volume available, because you're hitting the "mass limit per ton" first, as opposed to bumping up against the volume limit first. Conversely, some lower density items (finished goods that are not especially compact, such as an atmospheric fighter jet) will "waste" a lot of volume simply due to how they're shaped in comparison to the "box" that they would be fit into.

This is how you can have an F-16 fighter jet need 53 tons of displacement capacity.

And just for anyone playing the home game, the empty weight of an F-6 divided by a volume defined by the dimensions of an F-16 (length, width, height) yields an overall average density (specific gravity) of 8.5/52.5 ≈ 0.1619 ... or slightly under 1/6th the density of water, when the F-16 is "safed" for transport ... because so much of the volume defined by the aircraft's maximum dimensions is filled with "mostly empty air" when transported aboard another craft.

I’d say real world shipping uses the volume or mass whichever limit is hit first.

Here is a density of commodity in kg per square meter table/

Densities of common Products

Densities of common products - Imperial and SI-units.

www.engineeringtoolbox.com

 

[Grav_Moped]

Sorry, but that does not make much sense unless the standard interstellar cargo is goose down.
1 td cargo hold being 14 m³ AND 1 tonne would leave most of the cargo hold very empty, most of the time, wasting a lot of expensive ship.

You can reasonably either say:
(with LBB2'77) that 1 Dton = 1 tonne (& hence not 14 m³) or,
(with LBB2'81) that 1 Dton = 14 m³ (& hence not 1 tonne).

The obvious disadvantage with the first assumption is that you have no handy way to scale deck plans.

Well, you do. The problem is that you then need to track the mass (and density) of every ship component, and start doing all your performance calculations in 1000kg tons. It's been done. It's a lot more work than it's worth...

 

[kilemall]

Well, you do. The problem is that you then need to track the mass (and density) of every ship component, and start doing all your performance calculations in 1000kg tons. It's been done. It's a lot more work than it's worth...

It’s the sort of thing that could be useful for dramatic throw stuff overboard license or limiting magic cargo hold of holding cash in shenanigans but not constant calculations.

 

[Grav_Moped]

It’s the sort of thing that could be useful for dramatic throw stuff overboard license or limiting magic cargo hold of holding cash in shenanigans but not constant calculations.

Yeah. That also has implications for deck plans. Just because artistic license lets you put all of the possible extra "within 20% of tonnage is ok" margin for your custom Far Trader design into the cargo hold to give it another 80 ea. 1.5m deck squares, doesn't mean you've actually added another 40 tons of payload capacity!

 

[AnotherDilbert]

Well, yes, that makes sense. It's just never explained that way in the rules.

Mongoose isn't very detailed, on purpose. And that is presumably why people actually play it, unlike MT and TNE...

'Only the fuel tank is 14m^3/ton' or something would be helpful, or counting space and mass separately like in the Military Vehicles supplement.

Sorry, I can't be bothered to lookup the reference in the MgT1 book, but:

MgT2 Core22, p149:
The size of a ship, its components and cargo is measured in displacement tons or simply tons. A displacement ton displaces a volume equal to one ton of liquid hydrogen. One displacement ton is roughly equal to 14 cubic metres.

In MgT, 1 Dton ≈ 14 m³, no mass specified.

In MT, where mass and volume where specified separately:

MT RM, p46:
TRADE DEFINITIONS
Several terms and concepts are used on the trade and commerce flowcharts.
Lot: A lot is a single shipment of goods. A lot is identified by its displacement in tons (one ton equals 13.5 kiloliters).

[1 Dton = 13.5 kilolitre = 13.5 m³]

MT RM, p85:
To compute the average weight of a full cargo hold, multiply the volume of the cargo hold in kiloliters by 1000 kg (one metric ton).

1 Dton cargo ≈ 14 m³ ≈ 14 tonnes ≈ 14 000 kg. That is the density of water; metals and stone are heavier, packing materials are lighter. Actual cargoes are probably somewhat lighter, so you can generally stuff the cargo hold full without problems.

That is within reason, unlike 1 ton ≈ 14 m³ ≈ 1000 kg.

As a comparison a basic TEU container is ~38 m³ and max 24 000 kg (or 30 500 kg), so up to about 600 - 800 kg per m³. A design limit of 1000 kg per m³ is a simple even number that is not too unrealistic...

We don't normally haul lead, of course, but most of our cargoes don't fill the cargo bay floor-to-ceiling, nor do they cover the floor. They stack neatly, with place to walk between them.

That is great in a warehouse, but not so great in a moving vehicle where the stacks can fall over. The cargo presumably has to be secured in three dimensions, like in an aircraft.

 

[AnotherDilbert]

Let's use a slightly remarkable contemporary item then ... an F-16 ... to illustrate the point.

• Length: 15.06m
• Width: 9.96m
• Height: 4.9m
• Empty Weight: 8,573 kg (because we're shipping one as cargo)

15.06 * 9.96 * 4.9 / 14 = 52.49916 displacement tons
Mass = 8.573 tons

That is of course not how aircraft are shipped, or even stored in a carrier hangar.

But, yes, that is how a 5000 kg Fixed Wing Aircraft from LBB3 needs a 5 Dton ≈ 70 m³ hangar aboard ship in LBB2.

 

[AnotherDilbert]

I’d say real world shipping uses the volume or mass whichever limit is hit first.

Yes, of course, both are design limitations.

Here is a density of commodity in kg per square meter table/

Densities of common Products

Densities of common products - Imperial and SI-units.

www.engineeringtoolbox.com

Popped popcorn seems to be below the 70 kg/m³ (≈1000 kg/14 m³) limit. Sawdust is too heavy.

Iron ore is ~2400 kg/m³ or 33 600 kg/Dton.


I would say 1000 kg/m³ (14 000 kg/Dton) is a more reasonable assumption than 70 kg/m³ (1000 kg/Dton) for generic cargo.

 

[AnotherDilbert]

And IIRC, one or more of the later rule sets gave an average mass of 10,000kg/Td.

I can't recall any such definition.

CT Striker, MT, TNE, and T4 used 1 tonne = 1000 kg per m³ (~14 000 kg/Dton) for cargo.

Drives and other machinery is heavier, but as ships have lots of nearly empty space (staterooms, fuel), so total mass of (unarmoured) ships tend to be about 10 000 kg/Dton.

E.g. from MT:

 

[mike wightman]

FF&S

Thrust requires both energy and reaction mass, and so additional
fuel tankage will be required as well. Thrust is measured in
tonnes. Spacecraft require (for the sake of simplicity) 10 tonnes
of thrust per displacement ton to achieve an acceleration of 1 G.
Spacecraft with a final mass of more than 1 5 times (rounding fractions
to the nearest whole number) their hull rate (in displacement tons)
should recalculate their acceleration based on the adual thrust-tomassratio,
dividing thrust (in tonnes) by mass (in tonnes) to determine
acceleration in Cs (round fractions down). Most spacecraft, however,
will mass less than 15 tonnes per displacement ton.

 

[Spinward Flow]

That is of course not how aircraft are shipped, or even stored in a carrier hangar.

Some aircraft have folding wings.
I purposely chose an aircraft type that DOESN'T HAVE FOLDING WINGS to illustrate the point for you in order to make the distinction obvious.

I wasn't expecting you to declare the example "invalid" because it didn't yield the answer you were seeking in order to replace it with something more to your liking.